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A finite one-plaquette SU(2) regulator

Example

Let $Γ$ be the boundary graph of a square with vertices $v_{0}, v_{1}, v_{2}, v_{3}$ and one plaquette, traversed in the order $v_{0} \to v_{1} \to v_{2} \to v_{3} \to v_{0}$ . Label its successive boundary edges $e_{1}, e_{2}, e_{3}, e_{4}$ ; choose their reference orientations to be $v_{0} \to v_{1}$ , $v_{1} \to v_{2}$ , $v_{3} \to v_{2}$ , and $v_{0} \to v_{3}$ , respectively. Use the compact matrix group $G = S U (2) = {(\begin{array}{cc} a & b \\ - \overline{b} & \overline{a} \end{array}) : a, b \in C, ∣ a ∣^{2} + ∣ b ∣^{2} = 1}$ in its defining two-dimensional representation. Assign a link variable $U_{j} \in G$ to each edge in its reference orientation; in the ordered boundary product, a reversed traversal contributes $U_{j}^{- 1}$ . Thus the plaquette holonomy is $U_{p} = U_{1} U_{2} U_{3}^{- 1} U_{4}^{- 1}$ . The group is the unit sphere $S^{3}$ in the $(a, b)$ coordinates; let $μ$ be its normalized round surface measure. For $β \geq 0$ , define the dimensionless action $S_{β} (U) = β (1 - \frac{1}{2} Re Tr U_{p})$ and partition function $Z_{β} = \int_{G^{4}} e^{- S_{β} (U)} d μ^{\otimes 4} (U)$ . This finite graph integral satisfies $e^{- 2 β} \leq Z_{β} \leq 1$ , so it is finite and strictly positive. It has only four group variables and defines no continuum fields, no Wightman or Osterwalder–Schrader reconstruction, and no Hamiltonian; it therefore does not solve the four-dimensional existence and mass-gap problem.

Example references

Arthur Jaffe and Edward Witten, “Quantum Yang-Mills Theory,” §§4–6, printed pp. 141–150

Arthur Jaffe, “Constructive Quantum Field Theory,” §8, PDF p. 14, “For the Millennium: Gauge Theory in Four Dimensions”

Facts & Assumptions

Given: Use the finite graph, group, reference orientations, boundary traversal convention, and $β \geq 0$ stated in the Example.

[A1]

Each matrix in $G$ has the displayed form with $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , so $Tr U = 2 Re a$ .

[A2]

The measure $μ$ is normalized round surface volume on the unit sphere in $C^{2}$ , and $μ^{\otimes 4}$ is the corresponding fourfold product probability measure.

[A3]

The Hilbert–Schmidt inner product on complex matrices is $⟨ X, Y ⟩_{H S} = Tr (X^{†} Y)$ .

Proof

1.1givenA1algebra

Products and inverses of matrices in $G$ remain in $G$ , so $U_{p} \in G$ . By [A1], $Re Tr U_{p} = 2 Re a_{p}$ for some $a_{p}$ with $∣ a_{p} ∣ \leq 1$ . Hence $- 2 \leq Re Tr U_{p} \leq 2$ , and the defined action satisfies $0 \leq S_{β} (U) \leq 2 β$ .

1.2givenA2A3algebra

Left and right multiplication by any fixed $V \in G$ preserve the Hilbert–Schmidt inner product [A3]: $Tr ((V U)^{†} (V U)) = Tr (U^{†} U)$ and $Tr ((U V)^{†} (U V)) = Tr (U^{†} U)$ by unitarity and cyclicity of trace. The tangent variation $(\dot{a}, \dot{b})$ maps to a matrix with squared Hilbert–Schmidt norm $2 (∣ \dot{a} ∣^{2} + ∣ \dot{b} ∣^{2})$ , so the displayed embedding induces twice the standard round metric on the $(a, b)$ sphere. These multiplications therefore preserve its round volume and normalization. Thus $μ$ is a left- and right-invariant probability measure, and the product measure in [A2] has total mass one.

2.1givenA1A2step 1.1step 1.2algebra

Exponentiating the action bounds gives $e^{- 2 β} \leq e^{- S_{β} (U)} \leq 1$ for every $U \in G^{4}$ . The integrand is continuous, hence measurable and integrable on the compact probability space $(G^{4}, μ^{\otimes 4})$ . Integrating the pointwise bounds yields $e^{- 2 β} \leq Z_{β} \leq 1$ , which proves finiteness and strict positivity. At $β = 0$ , the integrand is identically one and $Z_{0} = 1$ .

3.1givenstep 2.1algebra∎

The defined object is one integral over the finite-dimensional space $G^{4}$ ; its definition contains no fields on $R^{4}$ , no limiting construction, no reconstructed Hilbert space, and no Hamiltonian whose spectrum could satisfy the required gap condition. It establishes only the finite-regulator integral asserted in the Example, not a continuum Yang–Mills theory or a mass gap.

Remarks

This finite construction is a regulated example related to Four-dimensional Yang–Mills existence and mass gap, not a logical approximation theorem or a supplier of the open conclusion.

A finite one-plaquette SU(2) regulator

Example

Example references

Facts & Assumptions

Proof

Remarks

Remarks references

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